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 Sep 30, 2019
edited by SoulSlayer615  Oct 23, 2019
 #1
avatar+2417 
+2

a) Hint: Plus it in

 

\(\sqrt{5^2+12^2}\), then solve. You should get an integer as an answer

 

b) To do this problem, you can square both sides.

 

\(x^2+2xy+y^2=x^2+y^2\)

 

Then cancel out x^2 and y ^2

 

\(2xy = 0\)

 

Then solve

 

\(x*y=0\).

 

So when x and y multiplied together equaling 0 is when it can satisfy the original equation.

 

Examples: (0, 0)  (10, 0)    (0, 5)

 

 

Values of x and y that DO NOT satisfy the original equation is when they DO NOT multiply to 0

 

Examples (1, 1) (2, 3) (6, 9)

 Sep 30, 2019
 #2
avatar+974 
+3

I just graphed it, you can find it out!

 Oct 1, 2019
 #3
avatar+2417 
+1

Yes, when graphed, it will always end up on the axis lines!

CalculatorUser  Oct 1, 2019
 #4
avatar+104962 
+1

x + y  =  sqrt (x^2 + y^2)      square both sides

 

x^2 + 2xy + y^2  =  x^2 + y^2

 

2xy  = 0

 

xy  =  0

 

Then.....this will be true as long as either x,y  = 0   or both = 0

 

 

Similarly

 

x + y  will not equal sqrt (x^2 + y^2)    whenever x and y are both non-zero

 

 

cool cool cool

 Oct 1, 2019

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